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Jerch
 Post subject: Lenny Conundrum 221
PostPosted: Fri Jul 06, 2007 2:21 am 
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A Two Hundred Dubloon Coin is unique in that the outer edge is gold while the inside is silver. Assuming both sides of the coin are identical, and a line tangent to the inner circle is 23 mm from edge to edge of the coin, and excluding the edge of the coin in the calculation, what is the surface area of gold on the coin, in square millimetres? Please round up to the nearest square millimetre.

Well...back to maths.

Anyway, I believe that they've assumed that the outer and inner 'ring/circle' are concentric.

All you really need is to use:
1. A pair of congruent triangles (either that, or the property of an isosceles triangle)
4. Pythagoras Theorem
5. Equation of the area of a circle
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chibis223
 Post subject: Re: Lenny Conundrum 221
PostPosted: Fri Jul 06, 2007 4:30 pm 
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Actually, all you REALLY need to know is:
Annulus: The area between two concentric circles
... and the ensuing propert(y/ies)
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Duke of Earl
 Post subject: Re: Lenny Conundrum 221
PostPosted: Fri Jul 06, 2007 5:27 pm 
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chibis223 wrote:
Actually, all you REALLY need to know is:
Annulus: The area between two concentric circles
... and the ensuing propert(y/ies)
That's so fundamental though... :p.

It would seem to me that you need to do some pixel measurements to determine a few numbers you need anyway. Or else I missed a key point.

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Regan
 Post subject: Re: Lenny Conundrum 221
PostPosted: Sat Jul 07, 2007 4:11 am 
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This question makes me groan only because it's been a few years since I've been in any math related classes and I could have easily answered this question if I were just a bit younger. Bah.

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draconis
 Post subject: Re: Lenny Conundrum 221
PostPosted: Sat Jul 07, 2007 10:34 am 
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I accidentally multiplied my answer by two, because I forgot I already doubled the values to consider both sides, so my final answer is definitely wrong. Pfft, what a waste of a LC I actually understood. :(

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AySz88
 Post subject: Re: Lenny Conundrum 221
PostPosted: Thu Jul 19, 2007 11:54 pm 
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Duke of Earl wrote:
It would seem to me that you need to do some pixel measurements to determine a few numbers you need anyway. Or else I missed a key point.
No need.
Let r = smaller radius, a = difference between smaller and larger radii.

Area = 2*pi*[(r+a)^2-r^2]

Right triangle formed by one large radius, one small radius, and the line:
11.5^2 + r^2 = (r+a)^2
--> r^2 = (r+a)^2 - 11.5^2

Plug into area formula:

Area = 2*pi*[(r+a)^2-((r+a)^2 - 11.5^2)] = 2*pi*[(r+a)^2 - (r+a)^2 + 11.5^2)]
= 2*pi*11.5^2
= 830.951 ~= 831

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